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Inequalities: Part 1

Prove that:

\(\displaystyle ab(a+b)+bc(b+c)+ac(a+c)\geq \sum_{cyclic}ab\sqrt{\frac{a}{b}}(b+c)(c+a)\)

where, \(\displaystyle \sum_{cyclic}ab\sqrt{\frac{a}{b}}(b+c)(c+a)=ab\sqrt{\frac{a}{b}}(b+c)(c+a)+bc\sqrt{\frac{b}{c}}(c+a)(a+b)+ca\sqrt{\frac{c}{a}}(a+b)(b+c)\)

So, prove that:

\(\displaystyle ab(a+b)+bc(b+c)+ac(a+c)\geq ab\sqrt{\frac{a}{b}}(b+c)(c+a)+bc\sqrt{\frac{b}{c}}(c+a)(a+b)+ca\sqrt{\frac{c}{a}}(a+b)(b+c)\)

Note by Saurabh Mallik
4 months, 4 weeks ago

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